Consider the series $$\sum_{n=1}^{\infty}{x^{2n}\over n^2+x^{2n}}$$

I want to show that the series is uniformly convergent in $[-1,1]$.

Theorem: A series of functions $\sum f_n$ will converge uniformly on $[a,b]$ if there exist a convergent $\sum M_n$ of positive numbers such that for all $x\in [a,b]$ $$|f_n(x)|\le M_n$$ for all $n$.


For $x\in [-1,1]\quad$ $|x^{2n}|\le 1\tag{1}$.


Reverse Triangle Inequality \begin{equation*} ||x|-|y||\le|x+y|. \end{equation*}

So $n^2-|x^{2n}|\le|n^2+x^{2n}|$ Therefore we have $n^2-1\le |x^{2n}+n^2|\tag{2}$

Using $(1)$ and $(2)$, we have $$|f_n(x)|\le {1\over n^2-1}$$

Let $a_n=1/n^2$, then ${M_n\over a_n}={n^2\over n^2-1}={1\over 1-{1\over n^2}}$ Therefor by Limit form comparison test, $\sum M_n$ converges.

The result follows.

Is this series uniformly convergent only in $[-1,1]$ and to what function it converges?

Edits: For $|x|>1$,

Divide by $|x|^{2n}$, we get $${1\over 1+{n^2\over |x|^{2n}}}$$. Since exponential grows faster than $n^2$ the expression goes to $1$ as $n\to \infty$

Therefore series diverges for $|x|>1$

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    $\begingroup$ Does the summand go to zero for $x>1$, as $n $ goes to infinity? $\endgroup$ – AnyAD Oct 23 '18 at 12:05
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    $\begingroup$ @AnyAD Thanks I understand it now. I have edited the post to show the calculation. $\endgroup$ – StammeringMathematician Oct 23 '18 at 12:24

Define $$f_n(x)={x^{2n}\over n^2+x^{2n}}$$In the interval $[-1,1]$ we have $0\le x^{2n}\le 1$ therefore $$|f_n(x)|=f_n(x)\le {1\over n^2+1}$$Now define $M_n={1\over n^2+1}$. Obviously the series $\sum M_n$ is convergent because$$\sum_{n=1}^{\infty}M_n=\sum_{n=1}^{\infty}{1\over n^2+1}\le \sum_{n=1}^{\infty}{1\over n^2}={\pi^2\over 6}$$and $$|f_n(x)|\le M_n$$which means that the sequence of $\{M_n\}$ fulfills the conditions of the theorem and using that theorem, we have proved what we wanted.

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  • $\begingroup$ I think you want $f_n(x)\le \frac{1}{n^2}.$ $\endgroup$ – zhw. Nov 30 '18 at 16:55
  • $\begingroup$ Also ${1\over n^2}$ is another possibility for $M_n$ since ${1\over n^2+1}<{1\over n^2}$ $\endgroup$ – Mostafa Ayaz Nov 30 '18 at 16:57
  • $\begingroup$ But why do you say $$f_n(x)\le {1\over n^2+1}?$$ $\endgroup$ – zhw. Nov 30 '18 at 17:01
  • $\begingroup$ I'm just looking for some $M_n$ so that the conditions of the theorem hold i.e. $$|f_n(x)|\le M_n$$since we have $$|f_n(x)|\le {1\over n^2}$$ we can arbitrarily choose $M_n={1\over n^2+1}$ $\endgroup$ – Mostafa Ayaz Nov 30 '18 at 17:05
  • $\begingroup$ But $1/(n^2+1)<1/n^2.$ $\endgroup$ – zhw. Nov 30 '18 at 17:08

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